Question 130
Let $a_1 \gt a_2 \gt \dots \gt a_n \gt 1$ be positive integers. Let $M$ denote the least common multiple of $a_1, a_2, \dots, a_n$. For a finite set of integers $X$, define
$$
f(X) = \min_{1 \le i \le n} \sum_{x \in X} \left\{ \frac{x}{a_i} \right\}.
$$
Here $\{u\} = u - \lfloor u \rfloor$ is the fractional part of the real number $u$. Put $f(\emptyset) = 0$. We say a set $X$ is *minimal*, if for any proper subset $Y \subsetneq X$, we have $f(Y) \lt f(X)$.
Prove that, if $X$ is a minimal finite set of integers and if $f(X) \ge \frac{2}{a_n}$, then
$$
|X| \le f(X) \cdot M.
$$
Here for a finite set $X$, we use $|X|$ to denote the number of elements in $X$.
$$
f(X) = \min_{1 \le i \le n} \sum_{x \in X} \left\{ \frac{x}{a_i} \right\}.
$$
Here $\{u\} = u - \lfloor u \rfloor$ is the fractional part of the real number $u$. Put $f(\emptyset) = 0$. We say a set $X$ is *minimal*, if for any proper subset $Y \subsetneq X$, we have $f(Y) \lt f(X)$.
Prove that, if $X$ is a minimal finite set of integers and if $f(X) \ge \frac{2}{a_n}$, then
$$
|X| \le f(X) \cdot M.
$$
Here for a finite set $X$, we use $|X|$ to denote the number of elements in $X$.